Question #e32b6

2 Answers
Sep 15, 2015

Answer:

#I_(3(aq))^(-) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)#

Explanation:

  • Oxidation:

#2S_2O_(3(aq))^(2–) -> S_4O_(6(aq))^(2–) + 2e^-#

  • Reduction: #I_(3(aq))^(–) + 2e^(-) -> 3 I_((aq))^(–)#

the overall reaction then:

  • Redox:

#I_(3(aq))^(–) + 2S_2O_(3(aq))^(2–) -> 3 I_((aq))^(–) + S_4O_(6(aq))^(2–)#

From the balanced equation, we can say that:

#(1)/"1" * n_(I_3^(–)) = (1)/"2" * n_(S_2O_3^(2–))#.

where the denominators 1 and 2 were taking from the corresponding coefficients of #I_3^–# and #S_2O_3^(2–)# respectively.

Now,

#[I_3^–]# . #V = (1)/"2" * [S_2O_3^(2–)] * V^'#

(since #n = C * V#), where

#V^' = "10.50 mL "# and #" "V = "15.00 mL"#

Therefore,

#[I_3^–] = (1)/"2" * [S_2O_3^(2–)] * (V')/"V"#

#[I_3^–] = (1)/"2" * "0.0500M" * (10.50 cancel("mL"))/(15.00 cancel("mL")) = "0.0175M"#

(rounded to 3 significant figures)

I would like to suggest the following video, that deals with a similar example in terms of stoichiometry of a redox reaction.

Sep 15, 2015

Answer:

#(a)" "I_(3(aq))^(-)+2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+3I_((aq))^-#

#(b)" "0.0175"mol/l"#

Explanation:

Thiosulfate ions give out electrons:

#2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+2e#

These are taken in by #I_3^-# which are reduced:

#I_(3(aq))^(-)+2erarr3I_((aq))^(-)#

Adding both sides gives:

#I_(3(aq))^(-)+2S_2O_(3(aq))^(2-)rarrS_4O_(6(aq))^(2-)+3I_((aq))^-#

(b).

#c=n/v#

#n=cxxv#

So no. moles #S_2O_3^(2-)=0.05xx0.0105=5.25xx10^(-4)#

From the equation we can see that the number of moles of #I_3^-# must be half of this:

no. moles #I_3^(-)=(5.25xx10^(-4))/(2)=2.625xx10^(-4)#

#c=n/v#

So #[I_(3(aq))^(-)]=(2.625xx10^(-4))/(0.015)=0.0175"mol/l"#